Question

Discuss the nature of bonding in \( [CoF_6]^{3-} \) and \( [Ni(CN)_4]^{2-} \) on the basis of valence bond theory (VBT) and find the value of magnetic moment in both.

OR
Explain the stereoisomerism in coordination compounds with suitable examples.

Hint:

Fix the metal oxidation state, then its d-count. Weak-field F^- keeps Co³⁺ (d⁶) high-spin (sp³d², 4 unpaired); strong-field CN^- pairs Ni²⁺ (d⁸) into dsp² square planar (0 unpaired). Use µ = √(n(n+2)). For the OR part recall geometrical (cis-trans, fac-mer) and optical (d-l) isomerism.

Answer 1

Option 1: Bonding by VBT and magnetic moments

The spin-only magnetic moment is \( \mu = \sqrt{n(n+2)} \) BM, where n is the number of unpaired electrons.

Step 1: Oxidation state and d-configuration in [CoF₆]^3-. Each F is -1 and the complex charge is -3, so cobalt is +3. Cobalt (Z = 27) is \( [Ar]3d^7 4s^2 \); removing 3 electrons gives Co³⁺ = \( 3d^6 \).

Step 2: Nature of ligand. F^- is a weak-field ligand, so it does NOT force pairing. The \( 3d^6 \) electrons stay high-spin as \( t_{2g}^4 e_g^2 \), keeping 4 unpaired electrons.

Step 3: Hybridisation and geometry. Since the inner 3d orbitals are occupied, cobalt uses outer orbitals: \( 4s, 4p, 4d \) giving sp³d² hybridisation. The complex is octahedral and outer-orbital (high-spin), hence paramagnetic.

Step 4: Magnetic moment. With n = 4, \( \mu = \sqrt{4(4+2)} = \sqrt{24} \).
\[\boxed{\mu_{[CoF_6]^{3-}} = 4.90\ \text{BM}}\]

Step 5: Oxidation state and d-configuration in [Ni(CN)₄]^2-. Each CN is -1 and the charge is -2, so nickel is +2. Nickel (Z = 28) is \( [Ar]3d^8 4s^2 \); removing 2 electrons gives Ni²⁺ = \( 3d^8 \).

Step 6: Nature of ligand and hybridisation. CN^- is a strong-field ligand, so it pairs up the \( 3d^8 \) electrons, freeing one 3d orbital. Nickel then uses \( 3d, 4s, 4p \) to give dsp² hybridisation and a square-planar, inner-orbital complex. All electrons are paired, so n = 0.

Step 7: Magnetic moment. With n = 0, \( \mu = \sqrt{0(0+2)} = 0 \).
\[\boxed{\mu_{[Ni(CN)_4]^{2-}} = 0\ \text{BM (diamagnetic)}}\]

Option 2: Stereoisomerism in coordination compounds

Stereoisomers have the same molecular formula and the same atom-to-atom bonds, but the atoms are arranged differently in space. Coordination compounds show two kinds:

Step 1: Geometrical (cis-trans) isomerism. It arises from different relative positions of ligands around the metal.
i) Square planar \( MA_2B_2 \): \( [Pt(NH_3)_2Cl_2] \) exists as a cis form (the two Cl at 90 degrees, adjacent) and a trans form (the two Cl at 180 degrees, opposite).
ii) Octahedral \( MA_4B_2 \): \( [Co(NH_3)_4Cl_2]^+ \) shows cis and trans forms.
iii) Octahedral \( MA_3B_3 \): \( [Co(NH_3)_3Cl_3] \) shows facial (fac, three like ligands on one face) and meridional (mer, three like ligands around a meridian) forms.

Step 2: Optical isomerism. It occurs when a complex and its mirror image are non-superimposable (chiral). The two forms, called d (dextro) and l (laevo), rotate plane-polarised light in opposite directions. Example: \( [Co(en)_3]^{3+} \) (en = ethylenediamine) exists as a pair of non-superimposable mirror images. The cis-\( [CoCl_2(en)_2]^+ \) ion is optically active, while its trans form is optically inactive because it has a plane of symmetry.

So coordination compounds display stereoisomerism as geometrical (cis-trans, fac-mer) and optical (d-l) isomerism.